Question

6^x+6^y=42
x +y =3 . find x and y ​

Answer 1

Answer:

$\sf{The \ values \ of \ x \ and \ y \ are \ 1 \ and \ 2 \ respectively}$

$\sf{or \ 2 \ and \ 1 \ respectively. }$

Given:

$\sf{6^{x}+6^{y}=42}$

$\sf{and \ x+y=3}$

To find:

$\sf{The \ value \ of \ x \ and \ y.}$

Solution:

$\sf{6^{x}+6^{y}=42...(1)}$

$\sf{x+y=3...(2)}$

$\sf{From \ equation(2), \ we \ get}$

$\sf{x=3-y}$

$\sf{Substitute \ x=3-y \ in \ equation(1), \ we \ get}$

$\sf{6^{3-y}+6^{6}=42}$

$\sf{\therefore{\dfrac{6^{3}}{6^{y}}+6^{y}=42}}$

$\sf{\therefore{\dfrac{216}{6^{y}}+6^{y}=42}}$

$\sf{Substitute \ 6^{y}=m}$

$\sf{\therefore{\dfrac{216}{m}+m=42}}$

$\sf{Multiply \ throughout \ by \ m, \ we \ get}$

$\sf{216+m^{2}-42m=0}$

$\sf{\therefore{m^{2}-42m+216=0}}$

$\sf{\therefore{m=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}}$

$\sf{\therefore{m=\dfrac{42\pm\sqrt{1764-4(216)(1)}}{2}}}$

$\sf{\therefore{m=\dfrac{42\pm\sqrt{900}}{2}}}$

$\sf{\therefore{m=\dfrac{42\pm30}{2}}}$

$\sf{\therefore{m=21\pm15}}$

$\sf{\therefore{m=36 \ or \ m=6}}$

$\sf{when \ m=36}$

$\sf{6^{y}=36}$

$\sf{\therefore{6^{y}=6^{2}}}$

$\sf{\therefore{y=2}}$

$\sf{Substitute \ y=2 \ in \ equation(1), \ we \ get}$

$\sf{x+2=3}$

$\sf{\therefore{x=1}}$

$\sf{when \ m=6}$

$\sf{6^{y}=6}$

$\sf{\therefore{y=1}}$

$\sf{Substitute \ y=1, \ in \ equation(1), \ we \ get}$

$\sf{x+1=3}$

$\sf{\therefore{x=2}}$

$\sf\purple{\tt{\therefore{The \ values \ of \ x \ and \ y \ are \ 1 \ and \ 2 \ respectively}}}$

$\sf\purple{\tt{or \ 2 \ and \ 1 \ respectively. }}$

Answer 2

Answer:

Your answer , x=1 and y=2

Step-by-step explanation:

Given x+y=3

6^x+6^y=42

Now , x+y=3

y=3-x

also , we can say , 6^x-6^(3-x)= 42

Multiplying with

6x

we get

6^2x+216=(42.6)^x

=6^2x + 42.6^x+216= 0

By middle term splitting , we get :

⇒6^2x−36.6x−6.6x+216=0

⇒6(6^x - 6 ) - 6(6^x-6) =0

= 6(6^x-6) = 6(6^x-6)

As we can see ,

L.H.S= R.H.S

.:. x = 1

and since,

y=3-x

Putting the value of x in equation

⇒y=3-1

y=2